We know that $\frac{4}{1+16x^2}=4-64x^2+1024x^4-16384x^6+...$ for $x\in\left(-\dfrac{1}{4},\dfrac{1}{4}\right)$. Using this fact, find the function that corresponds to the following series. $ 4x-\frac{{64}}{3}x^3+\frac{1024}{5}x^5-\frac{16384}{7}x^7+...$ Choose 1 answer: Choose 1 answer: (Choice A) A $\arctan(4x)$ (Choice B) B $\arctan(-4x)$ (Choice C) C $-\arctan(4x)$ (Choice D) D $\ln(-4x)$ (Choice E) E $-\ln(4x)$ (Choice F) F $\ln(4x)$
First, note that the derivative of $ 4x-\frac{{64}}{3}x^3+\frac{1024}{5}x^5-\frac{16384}{7}x^7+...$ is $ 4-64x^2+1024x^4-16384x^6+...=\frac{4}{1+16x^2}$ Then $\int\big({4-64x^2+1024x^4-16384x^6+...}\big)dx=\int\left(\frac{4}{1+16x^2}\right)dx$ We rewrite the integral on the right-hand side. $ \int\left(\frac{4}{1+16x^2}\right)dx=\int\left(\frac{4}{1+(4x)^2}\right)dx$ This gives us $\int{\frac{4}{1+(4x)^2}}~dx=\int\big({4-64x^2+1024x^4-16384x^6+...}\big)dx$ That is, $\arctan(4x) + C=4x-\frac{{64}}{3}x^3+\frac{1024}{5}x^5-\frac{16384}{7}x^7+...$ Now let $x=0\,$ ; since $\arctan(0)=0$, we know that $C=0$. Hence, $ 4x-\frac{{64}}{3}x^3+\frac{1024}{5}x^5-\frac{16384}{7}x^7+...=\arctan(4x)$